\(4\sqrt{x}=8\)

x=4

a) 16x=64

x=4

câu b và c là bình phương 2 vế

am-gm cái VT(đánh giá từ TBN sang TBC) 

\(ĐKXĐ:x\ge\frac{1}{2}\)

Áp dụng BĐT AM - GM cho các số dương ta có :

\(\sqrt{2x-1}=\sqrt{1.\left(2x-1\right)}\le\frac{1+2x-1}{2}=x\)

\(\sqrt[4]{4x-3}=\sqrt[4]{1.1.1.\left(4x-3\right)}\le\frac{1+1+1+4x-3}{4}=x\)

\(\sqrt[6]{6x-5}=\sqrt[6]{1.1.1.1.1.\left(6x-5\right)}\le\frac{1+1+1+1+1+6x-5}{6}=x\)

\(\Rightarrow\sqrt{2x-1}+\sqrt[4]{4x-3}+\sqrt[6]{6x-5}\le3x\)

Dấu "=" xảy ra \(\Leftrightarrow x=1\) ( Thỏa mãn ĐKXĐ )

Vậy pt có nghiệm duy nhất \(x=1\)

Hummmm

Dạ em không biết ạ,tại vì em mới học lớp 4 ạ,em xin lỗi ạ

pt<=>\(\sqrt{\left(x+6\right)^3}+\sqrt{x+6}=\left(x^2+4x\right)^3+x^2+4x\)

đặt\(\sqrt{x+6}=a;x^2+4x=b\)

A, đk tự tìm

\(\sqrt{x^2+4x+3}=x-2\)

\(\Leftrightarrow x^2+4x+3-x^2+4x-4=0\)

\(\Leftrightarrow8x-1=0\)

\(\Leftrightarrow x=\frac{1}{8}\)

B, đk tự tìm

\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\frac{4}{3}\sqrt{9\left(x+5\right)}\)=6

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow\sqrt{x+5}\left(2-3+4\right)=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow\sqrt{x+5}=2\)

\(\Leftrightarrow x+5=4\)

\(\Leftrightarrow x=-1\)

\(ĐKXĐ:x\ge-\frac{2}{3}\)

Ta có : \(4x^2+6x+1=4\sqrt{6x+4}\)

\(\Leftrightarrow4x^2+6x+1+6x+4+4=6x+4+4\sqrt{6x+4}+4\)

\(\Leftrightarrow4x^2+12x+9=\left(\sqrt{6x+4}\right)^2+2.\sqrt{6x+4}.2+2^2\)

\(\Leftrightarrow\left(2x+3\right)^2=\left(\sqrt{6x+4}+2\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x+3=\sqrt{6x+4}+2\left(1\right)\\2x+3=-\sqrt{6x+4}-2\left(2\right)\end{cases}}\)

+) Pt (1) \(\Leftrightarrow\sqrt{6x+4}=2x+1\)

\(\Leftrightarrow\hept{\begin{cases}5x+4=4x^2+4x+1\\x\ge-\frac{1}{2}\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)\left(4x+3\right)=0\\x\ge-\frac{1}{2}\end{cases}}\) \(\Leftrightarrow x=1\) ( Thỏa mãn )

+) Pt (2) \(\Leftrightarrow\sqrt{6x+4}=-2x-5\)

\(\Leftrightarrow\hept{\begin{cases}6x+4=\left(-2x-5\right)^2\\x\le-\frac{5}{2}\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}6x+4=4x^2+25+20x\\x\le-\frac{5}{2}\end{cases}}\) ( Vô nghiệm )

Vậy phương trình đã cho có nghiệm duy nhất \(x=1\)

\(\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1\)

\(pt\Leftrightarrow\sqrt{x^3+1}=\frac{2x^3+x^2+1}{4x-1}\)

\(\Leftrightarrow\sqrt{x^3+1}-\left(x+1\right)=\frac{2x^3+x^2+1}{4x-1}-\left(x+1\right)\)

\(\Leftrightarrow\frac{x^3+1-\left(x+1\right)^2}{\sqrt{x^3+1}+x+1}=\frac{2x^3-3x^2-3x+2}{4x-1}\)

\(\Leftrightarrow\frac{x^3-x^2-2x}{\sqrt{x^3+1}+x+1}-\frac{2x^3-3x^2-3x+2}{4x-1}=0\)

\(\Leftrightarrow\frac{x\left(x-2\right)\left(x+1\right)}{\sqrt{x^3+1}+x+1}-\frac{\left(x+1\right)\left(x-2\right)\left(2x-1\right)}{4x-1}=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(\frac{x}{\sqrt{x^3+1}+x+1}-\frac{2x-1}{4x-1}\right)=0\)

Suy ra x=2;x=-1 còn 1 nghiệm nữa xấu quá t gg :v

d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)

ĐK:\(x\ge-3\)

\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)

\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)

\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)

\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)

d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)

ĐK:\(x\ge-3\)

\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)

\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)

\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)

\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)